(i) For position coordinates p(x, y), Â the path of horizontal projectile is parabolic.
(ii) Â The time to reach the ground is 0.96 s.
(iii) The distance from the cliff of the ball on hitting the ground is 9.6 m.
(iv) The velocity with which the ball hits the ground is 9.4 m/s.
(v) The direction of the ball to the horizontal just before it hits the
ground is 90 degrees.
When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion. The object which follows the projectile motion.
Given is a ball is thrown forward horizontally from the top of a cliff
with a velocity of 10m/s. Â The height of the cliff above the ground is 4.5m.
(i) The equation formed by the ball is at point (x,y) given by
y =yo + Voy x t - 1/2gx²
The equation is in the form of parabola.
Thus, the ball follows parabolic path.
(ii) The time to reach the ground is
t = √2h/g
t =  √(2x 4.5)/9.81
t= 0.96 s
iii) Horizontal distance traveled
R = u x t
R = 10 x 0.96
R = 9.6 m
Thus, the distance from the cliff of the ball on hitting the ground is 9.6 m.
(iv) The velocity with which the ball hits the ground.
v = √2gh
Plug the values, we get
v = √(2 x 9.81 x 4.5)
v = 9.4 m/s
(v) The direction of the ball to the horizontal will be 90 degrees just before it hits the ground.
Learn more about projectile.
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